4 ms·

This kills it for me too. I rely on pypy, and I just can't see myself sacrificing my productivity by using Six. Also, Python 3 neglected to fix one of the most

• rspeer 11y ago
  Because nobody's made a Python Enhancement Proposal for what should happen instead.
• makecheck 11y ago
  It may be an easy mistake to make but using None is more consistent and efficient to detect. You can use "if x is None:" for any type; whereas, a function may or may not want to allow empty strings (or even empty lists) and it is easy to use a None check to decide when to apply the desired default.
• nbadg 11y ago
  Are you referring persistence of mutable default args? Which then potentially leads to bugs when you treat it as a new variable on each function call? This is intentional behavior. It's the result of one-time evaluation of default args, which is important for memoization. It's also really useful in combination with late-binding closures. For example, using a lambda as a generator function: def create_multipliers(): return [lambda x : i * x for i in range(5)] This doesn't work, you'll get all 8's. Instead you need to: def create_multipliers(): return [lambda x, i=i : i * x for i in range(5)] Late-binding closures and memoization strategies are pretty core language features; I wouldn't expect them to change (and many python devs would be pretty pissed if they did). Yes, this can be confusing with mutable default objects, but the alternative is to have disparate behavior depending on the mutability of the defaults, which would be an utter catastrophe. http://docs.python-guide.org/en/latest/writing/gotchas/ http://docs.python-guide.org/en/latest/writing/gotchas/
  • alextgordon 11y ago
    Other languages seem to get it right. Even C++! #include <vector> static int foo(std::vector<int> x = {}) { x.push_back(10); return x.size(); } int main (int argc, char const *argv[]) { printf("%d\n", foo()); // 1 printf("%d\n", foo()); // still 1 return 0; } def foo(x=[]): x.append(10) return len(x) print(foo()) # 1 print(foo()) # 2? wtf?
    • eiopa 11y ago
      ES6 got this right too. This design decision makes me die a little every time I have to do: def foo(x=None): if not x: x = [] ...
      • dajonker 11y ago
        That will get you into trouble as well, please use: if x is None: x = [] ... etc.
        • switch007 11y ago
          Or even "prettier": x = [] if x is None else x
          • knite 11y ago
            I'm partial to: x = x or []
            • acveilleux 11y ago
              Only works when x is truthy for all legal x values, excluding x=0.
    • deleted 11y ago
      [deleted]
    • 11y ago
• throwaway91919 11y ago
  Can you (or someone) clarify what problem you're talking about? This seems to work as expected for me: def foo(x=[]): sum = 0 for a in x: sum += a return sum print foo([3,5]) # 8 print foo(x=[3]) # 3 print foo([]) # 0 print foo() # 0 Is it the fact that kwargs are required to have defaults? Or what?
  • raverbashing 11y ago
    Nah, do it like this def foo(i,x=[]): x.append(i) return x foo(1) foo(2) See what happens
  • saltylicorice 11y ago
    The problem is that the default argument is mutable. def f(a=[]): a.append('v') print a f(['ok']) # ['ok', 'v'] f() # ['v'] f() # ['v', 'v']
    • throwaway91919 11y ago
      Ah, thanks.
• deleted 11y ago
  [deleted]